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The minimum value of $3x + 4y$ subject to the condition

$$x^2 y^3 = 6$$

and $x$ and $y$ are positive .

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3 Answers 3

up vote 9 down vote accepted

Write

$$3x = \frac{3x}{2} + \frac{3x}{2}$$

$$4y = \frac{4y}{3} + \frac{4y}{3} + \frac{4y}{3} $$

and use $\text{AM} \ge \text{GM}$.

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This solution uses elementary calculus, which you may not be familiar with. Given that $x,y>0$ we can solve for $x$ in terms of $y$ as $x=\sqrt{6/y^3}$, so the problem becomes minimizing $3\sqrt{6/y^3}+4y$ with $y>0$. This can be done by taking the derivative and setting it equal to $0$: $$0=\frac{d}{dy}\left(3\sqrt{6/y^3}+4y\right)=\frac{3}{2\sqrt{6/y^3}}\frac{-18}{y^4}+4=\frac{-27}{\sqrt{6y^5}}+4$$ and solving, which gives us $27^2=4^2\times 6y^5$ so $y=\sqrt[5]{\frac{27^2}{4^2\times 6}}=\frac{3}{2}$ is the value of $y$ which minimizes the expression. From this you can calculate $x$ and the minimum value of the expression $3x+4y$.

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$$3x+4y = 2(\frac{3x}{2})+3(\frac{4y}{3})$$

According to weighted arithmetic mean and weighted geometric mean inequality,

$$\left(\frac{m_1x_1 + m_2 x_2 + m_3 x_3 +...+ m_n x_n}{m_1+m_2+m_3....m_n}\right)^{m_1m_1m_2m_3.....m_n} \ge x_1^{m_1}x_2^{m_2}x_3^{m_3}x_4^{m_4}....x_n^{m_n}$$

$$\frac{2(\frac{3x}{2})+3(\frac{4y}{3})}{2+3} \ge[[\frac{3x}{2}]^2[\frac{4y}{3}]^3]^\frac{1}{2+3}$$

$$\frac{3x+4y}{5} \ge[\frac{16x^2y^3}{3}]^\frac{1}{5}$$

$x^2y^3=6$

$$\frac{3x+4y}{5} \ge{32}^{1/5}$$

$$3x+4y \ge 10$$

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Can anyone suggest some way to make the exponent of the right hand side of the equation in the gray portion , more readable? –  Tomarinator Mar 18 '12 at 14:42
    
@Aryabhata , the exponents on the R.H.S of that equation aren't visible now. I mean $\frac{1}{m_1m_2...}$ –  Tomarinator Mar 18 '12 at 17:50
    
They have been moved to the LHS. –  Aryabhata Mar 18 '12 at 17:57
    
Oh , ok how did I not see that before?, anyways thanks for the edit. –  Tomarinator Mar 18 '12 at 18:03

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